NCERT Solutions for Class 10 Maths Chapter 15 Probability Exercise 15.2

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NCERT Solutions for Class 10 Maths Chapter 15 Probability Exercise 15.2 Solutions are based on the following concepts:

Occurrence of an event
Impossible Events
Sure Event

Download PDF NCERT Solutions for Class 10 Maths Chapter 15 Probability Exercise 15.2

Check out NCERT Solutions for Class 10 Maths Chapter 15 Probability Exercise 15.2 Solutions

Exercise Solutions of Class 10 Maths Chapter 15 Probability

Also check other Exercise Solutions of Class 10 Maths Chapter 15 Probability

NCERT Solutions for Class 10 Maths Chapter 15 Probability Exercise 15.1 Solutions

Also Check:

Probability Chapter Related Topics
Theoretical Probability	Equally Likely Events	Empirical Probability
Events in Probability	Complimentary Events	A Theoretical Approach to Probability
P Value	Experimental Probability	Elementary Event
Sum of Probabilities	Geometric Probability	Chance and Probability
Empirical Probability Formula	Poisson Distribution Formula	Standard Normal Distribution

Also check:

Math Study Guides
Maths Important Formulas	Maths MCQs	Comparison Articles in Maths
Difference Between Topics in Maths	Math Study Material	Arithmetic
Class 10 Maths Notes	Permutations and Combinations	Mensuration
Trigonometry	Number Systems	Geometry
Math Formula	NCERT Solutions for Class 10 Maths	Algebra

CBSE X Related Questions

1.
In the given figure, PA and PB are tangents to a circle centred at O. If \( \angle AOB = 130^\circ \), then \( \angle APB \) is equal to :
- \( 130^\circ \)
- \( 50^\circ \)
- \( 120^\circ \)
- \( 90^\circ \)
View Solution
2.
D is the mid-point of side BC of \(\triangle ABC\). CE and BF intersect at O, a point on AD. AD is produced to G such that \(OD = DG\). Prove that OBGC is a parallelogram.
View Solution
3.
In a class test, the sum of Anamika's marks obtained in Maths and Science is 30. Had she got 2 marks more in Maths and 3 marks less in Science, the product of the marks would have been 210. Find the marks she got in the two subjects.
View Solution
4.
The HCF of 960 and 432 is :
- 48
- 54
- 72
- 36
View Solution
5.
PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If \(OP = 13\) cm, then find the length AB and PA.
View Solution
6.
Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)
View Solution

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NCERT Solutions for Class 10 Maths Chapter 15 Probability Exercise 15.2

Exercise Solutions of Class 10 Maths Chapter 15 Probability

CBSE X Related Questions

1.
In the given figure, PA and PB are tangents to a circle centred at O. If \( \angle AOB = 130^\circ \), then \( \angle APB \) is equal to :

2.
D is the mid-point of side BC of \(\triangle ABC\). CE and BF intersect at O, a point on AD. AD is produced to G such that \(OD = DG\). Prove that OBGC is a parallelogram.

3.
In a class test, the sum of Anamika's marks obtained in Maths and Science is 30. Had she got 2 marks more in Maths and 3 marks less in Science, the product of the marks would have been 210. Find the marks she got in the two subjects.

4.
The HCF of 960 and 432 is :

5.
PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If \(OP = 13\) cm, then find the length AB and PA.

6.
Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)

Similar Mathematics Concepts

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NCERT Solutions for Class 10 Maths Chapter 15 Probability Exercise 15.2

Exercise Solutions of Class 10 Maths Chapter 15 Probability

CBSE X Related Questions

1.In the given figure, PA and PB are tangents to a circle centred at O. If \( \angle AOB = 130^\circ \), then \( \angle APB \) is equal to :

2.D is the mid-point of side BC of \(\triangle ABC\). CE and BF intersect at O, a point on AD. AD is produced to G such that \(OD = DG\). Prove that OBGC is a parallelogram.

3.In a class test, the sum of Anamika's marks obtained in Maths and Science is 30. Had she got 2 marks more in Maths and 3 marks less in Science, the product of the marks would have been 210. Find the marks she got in the two subjects.

4.The HCF of 960 and 432 is :

5.PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If \(OP = 13\) cm, then find the length AB and PA.

6.Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
In the given figure, PA and PB are tangents to a circle centred at O. If \( \angle AOB = 130^\circ \), then \( \angle APB \) is equal to :

2.
D is the mid-point of side BC of \(\triangle ABC\). CE and BF intersect at O, a point on AD. AD is produced to G such that \(OD = DG\). Prove that OBGC is a parallelogram.

3.
In a class test, the sum of Anamika's marks obtained in Maths and Science is 30. Had she got 2 marks more in Maths and 3 marks less in Science, the product of the marks would have been 210. Find the marks she got in the two subjects.

4.
The HCF of 960 and 432 is :

5.
PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If \(OP = 13\) cm, then find the length AB and PA.

6.
Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)